Chemistry - Atomic Structure Question with Solution | TestHub

$\begin{array}{l}\text{IE}\text{=}\frac{{\text{Z}}^{\text{2}}}{{\text{n}}^{\text{2}}}\text{×}\text{13}\text{.6}\text{eV}\hfill \end{array}$…(i)
$\frac{{\text{IE}}_{\text{1}}}{{\text{IE}}_{\text{2}}}\text{=}\frac{{\text{Z}}_{\text{1}}{}^{\text{2}}}{{\text{n}}_{\text{1}}{}^{\text{2}}}\text{×}\frac{{\text{n}}_{\text{2}}{}^{\text{2}}}{{\text{Z}}_{\text{2}}{}^{\text{2}}}$…(ii)
Given${\text{IE}}_{\text{1}}\text{=}\text{19}\text{.6}\text{×}{\text{10}}^{-\text{18}}\text{,}{\text{Z}}_{\text{1}}\text{=}\text{2,}$
${\text{n}}_{\text{1}}\text{=}\text{1,}{\text{Z}}_{\text{2}}\text{=}\text{3}\text{and}{\text{n}}_{\text{2}}\text{=1}$.
Substituting these values in equation (ii)
$\frac{\text{19}\text{.6}\text{×}{\text{10}}^{-\text{18}}}{{\text{IE}}_{\text{2}}}\text{=}\frac{\text{4}}{\text{1}}\text{×}\frac{\text{1}}{\text{9}}$
Or${\text{IE}}_{\text{2}}\text{=}\text{19}\text{.6}\text{×}{\text{10}}^{-\text{18}}\text{×}\frac{\text{9}}{\text{4}}$
$\text{=}\text{4}\text{.41}\text{×}{\text{10}}^{-\text{17}}\text{J/atom}$